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math/9904079 | As follows from REF below, MATH. Therefore, it suffices to show that the correspondence MATH is injective map of MATH to MATH. The injectivity is manifest, so we only have to check that the image is MATH-invariant. Clearly, MATH. Therefore, it suffices to verify that MATH for every simple root MATH. This is subject to a direct check with the help of the multiplication table in MATH. |
math/9904079 | Since the dimensions of both modules are equal, it suffices to show that the natural homomorphism MATH is surjective, that is, the module generated by MATH coinsides with the whole module. The module in the right hand side has a natural filtration induced by filtrations of the modules MATH and MATH. Let the MATH be a basisi of MATH and MATH a basisi of MATH. Consider the module MATH generated by MATH, that is, by the elements of filtration zero and the element MATH . We have MATH . Since each summand is of filtration MATH, they belong to MATH by inductive hypothesis; hence, so does MATH. |
math/9904079 | Let MATH be any positive integer. Consider the map MATH such that MATH. Clearly, MATH is a MATH-module homomorphism because it is induced by projections of MATH and MATH onto their even parts. Take MATH and MATH from REF and consider the map MATH . Clearly, MATH is a MATH-module homomorphism. Let us consider the restrictions of the maps MATH and MATH onto MATH and MATH, respectively. Clearly, MATH sends the first of these spaces into the second one, whereas by REF MATH sends the second of these spaces into the first one. REF implies that, as MATH-modules, the spaces MATH and MATH have simple spectra. Let MATH while MATH correspond to a typical diagram MATH and both are nonzero, that is, MATH and MATH. Then the simplicity of the spectrum and REF imply that MATH and MATH are, up to a constant factor, mutually inverse isomorphisms of the modules MATH and MATH. Let MATH be the decomposition of the column stabilisor MATH of the tableau MATH with respect to the left cosets relative the product of the column stabilisors of MATH and MATH and same of the row stabilisor MATH. Then MATH . It is easy to verify that MATH . The vector MATH belongs to a typical module, is a highest one with respect to MATH and nonzero, where MATH is the set of strictly upper-triangular matrices with respect to the fixed basis of MATH. But REF implies that MATH is also highest with respect to MATH and lies in the same module. This shows that MATH . Further, from REF it follows that the vector MATH is MATH-invariant because MATH is MATH-invariant. We make use of the fact that MATH to decuce that MATH . Further on, MATH where MATH runs over all the sequences of length MATH composed from the integers REF to MATH. But, as is not difficult to see, MATH hence, MATH . Therefore, MATH which proves the theorem. |
math/9904079 | It is easy to verify that MATH, which immediately implies REF . REF is similarly proved. |
math/9904079 | For NAME tableaux MATH and MATH we have MATH . The dimension of this space is equal to either REF or REF. It is equal to REF only if MATH or both of them contain a MATH rectangle and MATH for MATH and MATH for MATH and any MATH. To prove the theorem, it suffices to show that for the above MATH and MATH the module MATH containes an invariant which can be expressed via the invariants listed in the theorem. By CITE such an invariant exists. Under the canonical homomorphism of the tensor algebra onto the symmetric one, the invariants of the form REF - iii turn into a system of generators. Theorem is proved. |
math/9904079 | To MATH there corresponds the identity operator MATH; hence, to MATH there corresponds the operator MATH and to MATH there corresponds the operator MATH; finally, to MATH there corresponds the operator MATH. Hence, MATH . The last equality follows from MATH. |
math/9904079 | Since for the representatives of the cosets of MATH we can take a collection of cycles, we may assume in REF that MATH . Hence, MATH. NAME on, MATH if and only if the last row of MATH for some MATH is, up to a permutation, a permutation of MATH. The set of marked tableau MATH is in one-to-one correspondence with the set of pairs MATH such that the last row of MATH is, up to a permutation, MATH. Indeed, from the pair MATH determine MATH, where MATH is the cycle that shifts the elements under the MATH-th marked one one cell up along the column and places the marked one at the bottom. If the marked element lies in the last row, we set MATH. And, the other way round, given MATH, we mark MATH, , MATH, where MATH is the last row of MATH. Hence, REF implies that MATH where MATH is a sign depending on MATH. Direct calculations of this sign lead us to REF . |
math/9904079 | Compare REF . |
math/9904079 | Compare REF . |
math/9904079 | Compare REF . |
math/9904079 | On MATH the group MATH and its subgroup MATH naturally act; namely, MATH permutes pairs MATH whereas MATH permutes inside each pair. Clearly, MATH. But, on the other hand, MATH, so MATH. Hence, in the decomposition of MATH only enter MATH for which MATH and their multiplicity is equal to MATH. But MATH so the multiplicity of MATH in MATH is equal to that of MATH in MATH. By CITE it is equal to REF if the lengths of all rows of MATH are even and REF otherwise. This proves REF. CASE: Consider now the natural map MATH. For the tableau MATH from the conditions of the theorem and the sequence MATH the vectors MATH form a basis of MATH. So the images of these vectors (which are exactly the MATH) form a basis of MATH. |
math/9904079 | By REF MATH . Hence, the kernel of homomorphism REF is equal to MATH. Let us prove that it is contained in the ideal generated by MATH, where MATH is a MATH rectangle. Let MATH and MATH the corresponding idempotent. Then MATH. Hence, MATH . Thus, MATH and we are done. |
math/9904079 | Set MATH . The map MATH is an isomorphism induced by the invariant bilinear form and MATH is a MATH-invariant. Let MATH be a rectangular MATH tableau as in REF and MATH a MATH-sequence such that after being filled each row MATH is of the form MATH. Denote by MATH the set of such sequences MATH. Then MATH where MATH, , MATH are the columns of the tableau MATH and where MATH whereas MATH. The element REF is a MATH-invariant; having applied to it the operator MATH from REF we get another MATH-invariant: MATH where MATH . For the collection MATH define the sequence MATH as follows: if MATH and MATH lie in the same row just strike them out, if MATH and MATH lie in distinct rows we strike them out and place their conjugates, MATH and MATH, in the last row in the same columns. The sequence MATH takes the form MATH. It is not difficult to verify that MATH . Therefore, MATH . The constant factor, the sign, can be, clearly, replaced with a REF. If MATH is of the above form, then in the last row for some values of MATH the pairs MATH are conjugate whereas all the remaining values of MATH are odd and pairwise distinct, call them MATH. Set MATH where MATH is the sign of the permutation. Then MATH and, therefore, MATH . To complete the proof, it suffices to calculate the number of pairs-MATH that give the sequence MATH which leads to REF . |
math/9904079 | Let MATH be a MATH-invariant which is not MATH-invariant. Let MATH depend on MATH even and MATH odd generic vectors MATH, , MATH. Then there exists a MATH such that MATH. Let MATH whereas MATH for MATH. Then MATH and MATH. But, on the other hand, MATH, hence, MATH. This means that MATH-invariants other than scalar products may only be of type MATH corresponding to a typical module. So, in the same vein as for MATH-invariants, we see that MATH if MATH is typical and its first MATH rows are of odd lengths whereas the remaining rows are of even lengths. If we do not consider the scalar products, then for the other (atypical) MATH there are no invariants. Under the canonical homomorphism MATH the module MATH turns into its copy and a basis of the first copy becomes a basis of the second one. This shows that if MATH is a MATH rectangle, then the polynomials MATH from the theorem constitute a basis of MATH, a subspace of MATH. For an arbitrary MATH containing a MATH rectangle we apply the same arguments as in the proof of REF . |
math/9904079 | Consider the two gradings of MATH: MATH and MATH . It is clear that MATH. If MATH is a MATH-invariant, then MATH (for MATH) and MATH for MATH. Therefore, MATH, where MATH. We similarly establish that MATH, where MATH. Hence, MATH. Taking into account the equality MATH we deduce that MATH and MATH for some MATH and MATH. Moreover, it is clear that MATH and MATH are MATH-invariants. Conversely, if MATH and MATH are MATH-invariants, then a direct check shows that MATH and MATH are MATH-invariants. |
math/9904079 | See REF. |
math/9904079 | From the theory of MATH-rings it follows that MATH, where the sum runs over the MATH of the above described form. One can easily verify that for the tableau as indicated in the formulation of the theorem and a MATH-standard sequence MATH the image MATH in MATH is nonzero. Hence, for a fixed tableau MATH the canonical map MATH performes an isomorphism of MATH with MATH. This implies the theorem. |
math/9904079 | Clearly, MATH where MATH is of the same form as stated in Theorem. Since REF is a MATH-module homomorphism, its kernel is MATH. The fact that this kernel is generated by the elements of the least degreee is proved by the same arguments as for MATH. |
math/9904079 | Clearly, MATH is the product of MATH factors. In each factor, select either MATH or MATH. In the first case, for MATH, set MATH and MATH in the second case set the other way round. We get a matrix with the properties desired. The sign is obtained after reordering of the sequence of the MATH: MATH . This is performed by induction: first, the pairs MATH are moved to the end in increasing order, this accrues the exponent of the sign with MATH, then we reorder the elements with indices lesser than MATH, which adds MATH, then the elements of the sequence MATH are rearranged into the sequence MATH which adds MATH to the exponent, and, finally, the elements MATH, , MATH are placed onto the end of the MATH-th row adding MATH. Besides, if MATH, then MATH enters MATH with a minus sign; this adds MATH. |
math/9904079 | Let MATH be a MATH rectangle filled in along columns. Set MATH. Denote by MATH the tensor obtained from MATH by replacing the elements occupying positions MATH, , MATH with numbers MATH, , MATH, respectively. Then MATH where MATH is any of the numbers of the positions occupied by MATH. If MATH, the product being ordered in order of increase of indices MATH, then MATH where MATH and MATH, where MATH. Assume that MATH, where MATH is the number of the first element in the MATH-th column of tableau MATH. We thus get MATH . By continuing the process we get MATH where MATH and MATH is the number of the first element in the MATH-th column and where MATH is equal to the sum of the numbers of the places occupied by REF's, and where MATH coinsides with MATH everywhere unless the MATH; then the MATH-th entry of MATH is occupied by MATH. If MATH and MATH are two sequences and MATH and MATH are two tableaux of the same form such that after filling MATH with the elements from MATH and MATH with the elements from MATH one gets geometrically identical pictures, then MATH implies MATH. Indeed, MATH . Therefore, if MATH, we have MATH because thanks to the fact that MATH only contains ``odd" elements MATH. Hence, MATH where MATH and where MATH coinsides with MATH everywhere unless where MATH, then the MATH-the position is occupied by MATH. We are done. |
math/9904079 | is similar to that of REF. |
math/9904086 | If MATH, then taking the stalks at MATH, we have an associated cosimplicial abelian group MATH, and a corresponding augmented complex. Clearly MATH unless MATH. Since the partially ordered subset MATH has a minimal element, one sees easily that the stalk complex at MATH is contractible (note that if MATH, and MATH is a MATH-simplex in the nerve of MATH), the stalks at MATH of MATH and MATH are naturally isomorphic, where MATH is the cone over MATH with vertex MATH. |
math/9904086 | We first claim that if MATH is an irreducible REF locally closed subset such that MATH is locally free, then MATH is a constant sheaf associated to a finitely generated abelian group, which is a quotient of MATH. Indeed, MATH corresponds to a representation of the fundamental group of MATH (with respect to any convenient base point), while MATH corresponds to the trivial representation. The sheaf map MATH is then a morphism of local systems, whose image MATH is clearly a trivial (that is, constant) local subsystem of MATH. Now let MATH be admissible for MATH. As observed above, MATH is constant for each MATH, and so MATH is also admissible for MATH. From REF , it follows that MATH is MATH-constructible for the NAME topology. From the beginning of the exact sequence MATH of REF (for MATH on MATH) we have inclusions MATH . We see at once from the definitions that MATH is (the direct image on MATH of) a constant sheaf on MATH, for each MATH, and the natural sheaf map MATH is injective. Since MATH is a constant sheaf, we also have that MATH is an isomorphism. Now from the commutative diagram MATH we deduce that MATH is an isomorphism, and that the natural map MATH is surjective. This implies that MATH is surjective as well, and that MATH is admissible for MATH. The exact sequence in REF is obtained from the resolution of REF for MATH. |
math/9904086 | We first note that, by an argument with mapping cones and cylinders (rotating the distinguished triangles in REF-diagram), we may assume that MATH without loss of generality. For such a MATH the analogous result for the cohomology diagram arising from a REF-diagram in the category of sheaves has been proved by NAME and NAME CITE. The proof of this lemma is entirely analogous: regarding the given REF-diagram as a (bounded) double complex of complexes, one considers the total complex, which is a REF-term exact sequence of complexes, say MATH . Regarding this again as a double complex, there is a spectral sequence MATH (the limit is in fact REF). Then the conclusion of the lemma is interpreted as giving two (equivalent) ways of computing the differential MATH. |
math/9904086 | Let MATH. Then by REF , there exists a NAME open cover MATH of MATH such that MATH where MATH . Therefore MATH as shown in the commutative diagram below (where MATH stands for any of the MATH) MATH . This finishes the proof of the lemma. |
math/9904086 | From REF there exists an exact sequence of abelian groups MATH . The natural surjective maps MATH (for MATH) are maps of pure NAME structures of weight one, which are quotients of MATH. Hence MATH is identified with the kernel of a morphism of pure NAME structures of weight REF, and hence itself supports a pure NAME structure of weight one. Also it is clear from the construction that the composition MATH is a direct sum of the natural quotient maps MATH, and hence is a morphism of NAME structures. Hence MATH is one as well. |
math/9904086 | Let MATH as above, and let MATH. We note that the natural map MATH is an injection (any section in the kernel must vanish in all stalks). This implies that the map MATH (which is a morphism of mixed NAME structures) is injective in the following commutative diagram (here MATH). MATH . We are done, because all the arrows in the above diagram are injections, and the vertical arrows (on the left and right borders), as well as the lower horizontal arrow, are morphisms of mixed NAME structures. |
math/9904086 | It is easy to see, from REF , that MATH . We will prove, using REF , that given any element MATH and any preimage MATH some non-zero (integral) multiple of MATH lies in MATH . This will prove the assertion of the lemma. Since MATH there exists a finite NAME open cover MATH of MATH such that MATH in MATH for all MATH . Let MATH denote any one of these MATH's and consider again the above commutative diagram with exact rows and columns. MATH . We wish to apply REF to this diagram; for this, we need to know that this diagram arises by applying the cohomology functor to a suitable MATH-diagram in the category of complexes of abelian groups. It is clear that the above diagram arises by applying the cohomology functor to the following MATH-diagram, where all the objects are in the (bounded below) derived category of sheaves of abelian groups on MATH, and the rows and columns are exact triangles; here MATH are suitable cones. MATH . Applying the functor MATH yields a MATH-diagram in the derived category of abelian groups. Using NAME resolutions, this MATH-diagram in the derived category is seen to be the image of a MATH-diagram where all the objects are complexes of abelian groups and the rows and columns are short exact sequences of complexes. Since the arguments are standard, we omit the details. Returning to our cohomology diagram, note that the relative cohomology sequences MATH and MATH are sequences in the category of mixed NAME structures by REF . Since MATH therefore MATH by REF . This implies, by REF (that is, strictness of morphisms of mixed NAME structures with respect to MATH) that we can choose MATH such that MATH . Let MATH for some MATH, where the last isomorphism is because MATH is non-singular; then MATH that is, MATH. So we can choose a preimage MATH of MATH to be zero. On the other hand, chasing the diagram the other way, we get MATH which lifts MATH, and MATH; now take a lift MATH of MATH. By REF , we know that MATH modulo the images of MATH and MATH . Therefore MATH . This proves that MATH comes from MATH . Since MATH has a finite cover by such open sets MATH we see that MATH . |
math/9904086 | Let MATH be any point and MATH . Let MATH and MATH, where MATH is the maximal ideal. For each MATH we have the restriction maps MATH . We claim that the kernel of MATH for each MATH is a MATH-vector space. To see this consider the short exact sequence of NAME sheaves MATH where MATH denotes the exponential map, which makes sense as MATH is nilpotent. Considering the associated cohomology sequence we get MATH which proves the kernel of MATH, for each MATH, is a MATH-vector space. For each MATH and for each MATH we have a commutative diagram MATH . Let MATH be the composition MATH . Then, MATH and MATH are both closed subgroups of MATH hence are compact (topological) groups. Since MATH is a MATH-vector space it follows that MATH as any continuous homomorphism from a compact group to a MATH-vector space is zero (note that MATH and MATH are isomorphic to the corresponding analytic groups, by GAGA, and hence from the exponential sequence carry natural topologies, such that the restriction homomorphisms are continuous). Passing to the inverse limit we have a commutative diagram MATH where MATH stands for the completion of MATH along MATH. By NAME 's Formal Function Theorem [H, Ch. III, REF ] and the fact that MATH is an isomorphism [H, Ch. II, REF ], we have that MATH is an injection. Thus it follows that MATH . We have from the definition of MATH that the stalk of MATH at MATH, MATH . By our analysis so far we have proved that the natural map MATH is an inclusion, and it clearly factors through the the subgroup MATH . By the results of NAME REF there exists a commutative triangle MATH . Note that MATH is a MATH-vector space. Now MATH induces a map MATH . Thus we have a diagram MATH . Since MATH is a mixed NAME structure with weights MATH and MATH (by REF , as MATH is a projective variety), MATH injects into MATH. Thus it is clear from the above diagram MATH and so MATH is also a MATH-vector space. Hence the composite MATH is injective, as MATH is a compact group, from its definition. Let MATH be the stalk of MATH at MATH . Let MATH be the saturation of MATH in MATH . The inclusion MATH induces a natural map with finite kernel MATH . In fact this map factors as MATH . The second map is an inclusion and MATH is the image of MATH in MATH . We thus have a commutative diagram with surjective and injective maps as follows where we identify MATH with MATH . MATH . Since it is clear from the diagram that MATH and MATH are both the image of MATH in MATH it follows that MATH . Therefore there exists a map MATH which is an isogeny. We had proved that the sheaf MATH was constructible, that is, constant with groups MATH over locally closed sets MATH and this data gives rise to a flasque resolution of MATH in the NAME site (by REF ). It is then clear that analogous results hold also for the sheaf MATH where MATH denotes the saturation of the sheaf MATH in MATH (which is a torsion-free sheaf). Thus the abelian varieties MATH are constant quotients of MATH over the strata MATH hence so are MATH . This proves that the sheaf MATH is a constructible sheaf on MATH for the NAME topology, with admissible family MATH and further (by REF ) MATH has a flasque resolution similar to MATH . Taking global sections of the flasque resolution of MATH, we get an exact sequence MATH . Also it is clear from the definitions that MATH. Applying MATH on all the terms, we obtain a complex MATH . This complex is exact on the left and has finite homology in the middle, since MATH is torsion-free. Similarly taking global sections of the flasque resolution of MATH we get an exact sequence MATH . There exists a commutative diagram MATH where the two vertical arrows are isomorphisms. Hence the dotted arrow MATH exists, and is an inclusion with finite cokernel. Define MATH . Clearly this is an abelian variety, and there is an isogeny MATH . Also, by construction, the natural map MATH clearly factors through the map MATH . Thus we have an isogeny MATH . We finally note that there exists an isogeny MATH since MATH . This finishes the proof that MATH and MATH are isogenous. |
math/9904086 | Using the short exact sequence of sheaves REF , we get the following commutative diagram, whose right column is exact, MATH . Here, we claim the dotted arrow MATH exists (and is also injective) because the composition MATH is zero. This is obvious as this map can be described in the following way: given the image in MATH of a NAME locally trivial cohomology class MATH, consider a line bundle MATH on MATH which represents it, then consider the line bundle restricted to open sets MATH (where MATH open) and take the NAME classes of these restrictions. These give a global section of MATH which is zero as the line bundle came from a locally trivial cohomology class on MATH . |
math/9904086 | Consider the following diagram with exact rows and columns MATH . The above diagram comes from the following MATH-diagram in the category of sheaves (where the bottom row defines MATH and MATH is the inclusion). MATH . Let MATH such that MATH both in MATH and MATH (that is, MATH). By a diagram chase as before we get elements MATH and MATH in the group MATH . Both MATH and MATH are well-defined in the quotient group MATH . Now note that MATH . Hence we get REF maps MATH and MATH from MATH where MATH denotes the class of MATH in the quotient MATH . We claim that these two maps are nothing but our previously defined maps MATH and MATH respectively. It is clear that the map MATH is equal to MATH . This is because we got MATH by first taking a lift of MATH say MATH in MATH then restricting to MATH to get MATH and finally taking the class MATH . This is exactly how MATH was defined, so MATH . It is also clear that MATH as the extension class map is defined exactly the same way as the map MATH . Now by REF , we have that MATH . This implies that MATH . |
math/9904088 | It will suffice to show the pullback vectorial extension is universal. Since MATH, any extension of MATH by a vector group MATH is pulled back from a unique extension of MATH by MATH. This extension of MATH is a pushout from the universal vectorial extension, so the same holds for the pullbacks to MATH. |
math/9904088 | There is a boundary map MATH . Define MATH. Let MATH be the torus with character group MATH. For MATH let MATH be the line bundle on MATH corresponding under the map REF. As a MATH-algebra MATH . The map MATH in REF comes from the above inclusion MATH . For MATH, (as is well known, compare CITE III REF), MATH has trivial cohomology in all degrees unless MATH. The proposition follows by taking cohomology of REF. |
math/9904088 | The MATH-algebra MATH is filtered, with MATH . With respect to the exact sequences MATH it suffices to show the boundary map MATH is injective. Composing on the right with the evident map, it suffices to show the maps MATH are injective. But MATH and the map in REF is the map MATH, which is injective. |
math/9904088 | Take MATH. |
math/9904088 | By REF on p. REF, the cohomology in the middle is a subgroup of the group of extensions MATH. (Note, MATH.) By the classification of commutative algebraic groups in characteristic MATH, this ext group vanishes (compare CITE, pp. REF). |
math/9904088 | Let MATH be an absolute lifting. Replacing MATH with MATH we may assume MATH vanishes. Then MATH vanishes in MATH. Let MATH be as in REF , so MATH. The previous lemma implies there exists MATH with MATH. Then MATH is the desired invariant absolute form. |
math/9904088 | Pullback on invariant relative forms is injective, because MATH is generated by the image of MATH. It follows by dimension count that the first arrow REF above is an isomorphism. For the absolute forms we may consider the diagram MATH . The left and right hand vertical arrows are shown to be isomorphisms in the proof of REF . Hence the isomorphism on invariant relative forms implies the isomorphism REF on invariant absolute forms. |
math/9904088 | This follows because MATH . The second arrow is surjective because we have a diagram MATH . The bottom row is not a priori exact, but MATH (because MATH acts by MATH on MATH.) The middle vertical arrow is onto for example, because the NAME group of the generic fibre of MATH is zero. Indeed, MATH is rationally split, and the kernel has trivial NAME group by hypothesis. (Since the function field of the generic fibre equals the function field of MATH, any divisor on MATH can be moved by rational equivalence to avoid the generic fibre, that is, to be a pullback from the base.) Finally, the right hand vertical arrow is injective because, after making a base change MATH, one can think of MATH and MATH as quotients of vector spaces by lattices, and the map on lattices is surjective MATH. |
math/9904088 | One has as in REF MATH . Now MATH . Exactness of the sequence in REF implies that there exists an element MATH with MATH. Take MATH. |
math/9904088 | This is immediate from the lemma. |
math/9904088 | The first step is to compute MATH for MATH. Let MATH be a finite dimensional MATH-vector space, and suppose MATH is a vectorial extension MATH . We know by REF that this sequence pulls back from an extension of MATH by MATH. Let MATH be the exact sequence of functions of filtration degree MATH as in REF , and let MATH be the boundary map in cohomology. We have MATH . One has a filtration MATH with MATH and MATH. The corresponding spectral sequence looks like MATH . Let MATH be as in REF , so MATH is the maximal quotient torus of MATH. REF identifies MATH with MATH, and the invariance condition might be looked at on MATH. Let us write MATH, where MATH consists of the regular functions which vanish at MATH. Then MATH. Thus if a class MATH is invariant, where MATH and the MATH are linearly independent over MATH, then MATH thus MATH and MATH . So for MATH, this implies that MATH and for MATH this implies that MATH . In short: MATH . Thus, it suffices to prove the lemma with MATH replaced by MATH, so we may assume the quotient torus MATH. Since in this case MATH, one sees that pullback under the multiplication by MATH map, MATH acts on MATH by multiplication by MATH. It follows that the spectral sequence REF degenerates at MATH. In particular the eigenspace where MATH acts by multiplication by MATH on MATH is MATH . Note that as a quotient of MATH the space MATH is clearly invariant. Conversely, let MATH be the diagonal. Since MATH it follows that for MATH we have MATH so necessarily MATH, proving REF. A similar argument on MATH proves REF. We remark here that MATH is in a natural way a subspace of the regular functions on MATH, and REF takes the quotient of this by MATH. This is because we have forced the rigidification condition in REF . We return to the proof of REF . The exact sequence MATH defines a map MATH . By REF , as a group extension of MATH, the group MATH is defined by a unique map from MATH to a vector space. We claim that this map is the dual of MATH. To see this, one identifies MATH and MATH. Then it is well know that the universal vectorextension on MATH is MATH inducing the universal vectorextension MATH on MATH where MATH for any subscheme MATH. The map of complexes MATH induces an isomorphism on MATH. Indeed, MATH sends the exact sequence MATH to the exact sequence MATH so one has just to see that MATH is an isomorphism. But MATH via the exponential map and the quasiisomorphism CITE MATH allows to conclude. Define a bundle MATH on MATH by pullback MATH . Because of the isomorphism MATH, the top row of REF pulls back uniquely from an extension of MATH by MATH. There is a unique vectorial extension MATH such that the above extension of vector bundles coincides with MATH . From this we get a diagram (defining MATH and MATH. Here MATH) MATH . We get a diagram with exact rows MATH . REF gives isomorphisms MATH . These are two of the desired arrows for the diagram in the proposition. The natural map on relative connections MATH is an isomorphism. We note the following facts: CASE: MATH. CASE: MATH. Indeed, as remarked in the proof of REF one has MATH. One checks that the kernel is generated by divisors of degree MATH supported on MATH. CASE: MATH . CASE: MATH (This is seen by noting MATH, so one has a homomorphism MATH such that MATH is an isomorphism on global units modulo constants. The assertion then reduces to the case MATH, which is easy.) We build a diagram MATH . Since the left and right hand arrows are isomorphisms, it follows that the central arrow is as well, proving the lemma. The assertions of the proposition follow easily from the lemma. |
math/9904088 | For example, in the absolute case, the curvature of a line bundle with invariant absolute connection on MATH is a section MATH satisfying MATH. It is easy to see that such a section lies in the subsheaf MATH. The isomorphism REF follows from REF and the fact that pullback to MATH of invariant forms is injective by REF. The case of relative forms is similar and is left for the reader. |
math/9904088 | Let MATH be a basis for MATH at MATH a point with multiplicity MATH in MATH, and let MATH be a local parameter at MATH on MATH. Write MATH . We must show MATH is regular at MATH. But integrability of MATH implies that MATH, from which the assertion is clear. |
math/9904088 | Our hypotheses imply MATH has dimension MATH. Consider the diagram MATH . We have MATH exterior tensor product on MATH. The NAME formula gives MATH . There is an action of the symmetric group MATH on the pair MATH . The resulting action on MATH is alternating because of the odd degree cohomology, so the invariants are precisely MATH. There is an evident map MATH . To show this map is an isomorphism, it suffices to remark that one has a trace map MATH . Because MATH, the existence of such a trace follows from the projection formula and the trace in NAME cohomology with constant coefficients. |
math/9904088 | Write MATH for the NAME sheaf associated to the presheaf MATH for any NAME sheaf MATH on MATH. For MATH locally free, MATH for MATH by purity. Duality theory gives (here MATH runs through nilpotent thickenings) MATH . We want to show that this map is an isomorphism, compatible with the connection, thus yielding a quasiisomorphism of complexes MATH . The problem is local, so we can assume MATH with MATH smooth of codimension MATH in MATH. Now MATH represents MATH in the derived category of NAME sheaves on MATH, and in the derived category we may write MATH . In this way, we reduce to verifying REF in the case MATH. So, suppose MATH in MATH. We have MATH as MATH . By CITE, since MATH has no singularity along MATH, one has MATH . Thus MATH is an isomorphism. |
math/9904088 | Let MATH correspond to a line bundle MATH of degree MATH, we consider the exact sequence MATH . Suppose first MATH. Then MATH, so any trivialization in MATH lifts to MATH, and the space of such liftings is a torseur under MATH, a vector space of dimension MATH. (Note this is an affine torseur, not a projective torseur.) If MATH, MATH has dimension MATH, and the image MATH has codimension MATH. |
math/9904088 | The isomorphism on the right in REF implies we must show MATH. The assumption MATH means we have a MATH action by translation on MATH, and minimality of MATH implies that the connection is nontrivial on the fibres. The fibration is NAME trivial, so the NAME spectral sequence for NAME cohomology reduces us to showing MATH where MATH is an everywhere non-zero, translation invariant, relative MATH-form on MATH. In other words, for MATH we must show MATH has trivial cohomology. This is straightforward. |
math/9904088 | Note MATH. Extending the top sequence in REF one step to the left and using the previous lemma gives the left isomorphism. We have already seen the isomorphism on the right. |
math/9904088 | Let us write MATH, where MATH is the cycle map. Then by definition, the trivialization of MATH associated to MATH depends only on MATH, or equivalently only on MATH. We have MATH where MATH is the pullback of MATH via the MATH-th projection MATH. Suppose for a moment that the divisor of MATH (viewed as a section of MATH) is reduced, MATH. Let MATH be the point corresponding to MATH. We have MATH and MATH. Since MATH is a projective bundle, there is a surjection on tangent spaces MATH . Since MATH is spanned by REF, to show MATH vanishes at MATH, it suffices to show MATH . This expression equals MATH . Each term on the right vanishes because MATH in MATH. The general case (MATH not necessarily reduced) follows from this by a specialization argument. We postpone until REF the proof that MATH doesn't vanish at any other point of MATH. |
math/9904088 | Recall REF the relative invariant forms on MATH are the MATH defined by the expression MATH . Write MATH where the nonvanishing condition comes from the requirement that the form restricted to MATH gives a trivialization along MATH (see REF). If we write REF MATH we get the table MATH . Note if we give MATH all weight MATH, then MATH will be homogeneous of weight MATH. Comparing REF, it follows that if we expand MATH in terms of the MATH, omitting MATH and MATH, we find for suitable MATH . Looking at the weights, we see that for MATH while MATH with neither MATH coefficient MATH. Now generators of MATH are of the form MATH where MATH is a monomial in the MATH and MATH is as in REF. Relations are MATH . Because of REF one can use these relations to eliminate MATH from MATH by downward induction on MATH, starting from MATH. We are left with the case MATH with MATH. In this case we can apply REF. If MATH, we get the relation MATH . Using this, we can get MATH. If MATH and MATH the relation becomes MATH . If MATH is not a positive integer, we can arrange MATH. On the other hand, we claim that if MATH, then MATH is the residue of MATH along MATH. Indeed REF shows that in this case, MATH, and MATH is then just the local parameter in the point MATH (see REF). Thus the quasiisomorphism MATH forces MATH not to lie in MATH. |
math/9904088 | We have seen in the proof of REF that MATH vanishes at a point in MATH. We must show it vanishes at at most one point. Let MATH be a point. Write MATH with respect to the coordinates MATH REF. Staring at REF, the conditions that MATH are seen to be (recall MATH) for MATH . For MATH one gets the same list but with the last line (coefficient of MATH) omitted. Finally, using MATH and MATH one gets MATH . Since MATH, REF admit a unique solution for the MATH. Since we know MATH vanishes at at least one point of MATH, this point must lie in MATH. |
math/9904088 | Since MATH is (relatively) closed on MATH, one can write MATH . Lifting to an absolute form forces MATH . Here the MATH are linearly independent in MATH. Using MATH modulo MATH and taking residues along MATH yields MATH. Then computing MATH yields MATH . It follows that MATH, so MATH. Taking MATH again shows MATH is closed if MATH is integrable. |
math/9904088 | We have seen REF that this point MATH is determined by the condition that MATH. Changing MATH by a closed form pulled back from MATH changes the NAME connection and the connection at MATH in the same way, so we can assume MATH, that is, MATH. Write MATH . Write MATH. Then MATH . We have MATH . Since MATH, we see from REF that MATH . Thus, it will suffice to relate MATH and MATH in MATH. Note that each monomial MATH involves MATH for only one value of MATH, and the weight of MATH is MATH (see the discussion after REF). Suppose first the weight of MATH is strictly less than MATH. Let MATH be maximal such that MATH appears in MATH. From REF it follows that MATH . Here MATH. Define MATH. Note the weight of MATH is MATH, so in MATH we have (compare REF) MATH where MATH is a sum of terms of weights MATH and terms of weight MATH only involving MATH. Note that MATH because MATH. In this way we reduce to the case MATH. Our assumption on the weight implies MATH, so MATH . Together with REF and MATH, this enables us to reduce to MATH. Suppose now the weight of MATH is MATH. If MATH we can use the above argument, except in the case MATH. Here there are two subcases. If MATH, the MATH in REF is MATH in the expansion MATH, so MATH . This completes the proof in this case because the connections MATH and MATH are isomorphic. If, on the other hand, MATH is odd and MATH, the monomial MATH and MATH contains the term MATH. Thus, from REF we conclude MATH. The lefthand identity in REF yields in this case MATH . In the statement of the lemma, we take MATH to be the product of the corresponding MATH. Suppose finally MATH. In this case MATH, so the corresponding MATH and by REF this term contributes nothing to MATH. Similarly, by REF there is no contribution to MATH. |
math/9904088 | First we collect some facts about MATH. We have MATH and MATH. It follows from REF of the paper that MATH does not appear in the expression for MATH and MATH only appears with constant coefficient. Restricting to MATH thus has the effect of surpressing the term in MATH and changing the coefficients MATH by a constant for MATH. Expressed in this way, it follows that the coefficient of MATH in MATH involves only monomials in MATH for the same MATH. Giving MATH and MATH both weight MATH, the terms in MATH all have weight MATH. It follows from REF that, writing MATH so MATH, we may write MATH . Here MATH. giving MATH and MATH weights MATH, all terms with first index MATH have weights MATH. All terms of the form MATH occur with nonzero coefficient. Notice that replacing MATH with MATH introduces monomials of lower degree, but these have weight MATH. It follows that MATH is the determinant of a matrix MATH where MATH is symmetric, MATH (respectively, MATH), and has the shape MATH with the entries MATH non-zero. Mod squares, MATH is MATH if MATH is even, and is given by MATH if MATH is odd. Writing MATH as just above REF with MATH we find MATH . (The sum on the right is over all MATH such that MATH.) |
math/9904088 | The strong NAME theorem identifies the determinant connections on MATH and MATH so we need only consider the connection on MATH. As well known, the NAME duality morphism MATH is compatible with the NAME connection, which is trivial on MATH. On the other hand, it is alternating, thus its determinant MATH fulfills MATH where MATH is the Pfaffian of the determinant of MATH, written in the basis MATH. Thus if MATH, one has MATH . Thus MATH and the determinant of the NAME connection is trivial. |
math/9904088 | Since MATH, where MATH is the ideal sheaf of MATH, MATH . Also one has MATH. Thus MATH . |
math/9904088 | The map MATH is surjective, and its kernel is the MATH-module generated by MATH where MATH is the connection with logarithmic poles along MATH with residue -REF. Let MATH be a local coordinate around MATH. Let MATH be a NAME covering of MATH, with MATH, and MATH. Assume MATH is the only zero or pole of MATH on MATH. Let MATH be a NAME representative of a class in MATH. Then MATH with MATH, MATH for MATH, MATH, MATH for MATH is a NAME representative of MATH. Thus considering MATH with MATH and MATH for MATH, the cocyle of REF is just the coboundary MATH . (Note MATH is invertible on MATH for MATH so the MATH-cochain is defined.) |
math/9904088 | Note that MATH so the complex in REF is indeed a quotient. From the diagram MATH one deduces that the left hand side of REF is isomorphic to MATH . The right hand side here is identified under the norm with MATH, which proves the second equality. The third one comes from the map MATH and the vanishing of MATH. Note that this cumbersome way of writing this cohomology allows to write local contribution of a class in this cohomology group. |
math/9904088 | The quasiisomorphism condition is local about each point of MATH, so we may assume our line bundles are MATH for some MATH and MATH. Choose local coordinates MATH near MATH and a NAME covering MATH of MATH such that MATH, MATH, MATH for MATH. Let us denote by MATH the class defined by the local trivialization MATH . Let MATH. Then MATH with MATH such that MATH and MATH . Suppose MATH. We drop the index MATH for convenience. One has MATH where the last term is a cocycle as in REF or the quotient complex REF. For MATH one replaces MATH by MATH, leaving MATH and MATH unchanged. By REF , one has MATH . Thus MATH . The nontrivial part in the last expression is computed in MATH . Computing using the residue at MATH we find the above difference is MATH . The verticality condition for the curvature reads MATH . In particular, MATH. The difference of the two products is therefore MATH, which vanishes in MATH. |
math/9904088 | Given the main result of CITE, and REF , the theorem is of course equivalent to MATH . Keeping the same notations as in the proof of REF , one has MATH and thus MATH . This lies in MATH and by REF , its trace factors through MATH. But the image of MATH in MATH is the relative NAME class MATH REF, thus the image of MATH in MATH is MATH, where MATH is the absolute NAME class of MATH. Indeed, MATH factors through MATH by NAME theory. On the other hand, if MATH is a cocyle for MATH, then MATH is a cocyle for MATH, and consequently, MATH. |
math/9904088 | One reduces easily to the case MATH is a single MATH-point. Let MATH be a NAME set containing MATH and MATH. Shrink MATH if necessary so there exists MATH with MATH and MATH. Let MATH so MATH. Shrinking the MATH if necessary, we can assume MATH, so MATH is represented by some cocycle MATH. Then MATH . On the other hand, by REF , MATH is represented by the image of the cocycle MATH . Write MATH with MATH regular on MATH. Since MATH extends to MATH, it is homologous to zero, so MATH . Now, since one obviously has MATH and the translation MATH is invariant, the second equality is a direct interpretation of the first one. |
math/9904088 | If all MATH this is just REF . So we assume that MATH in the sequel. Then as in the proof of REF , replacing MATH by MATH changes MATH to MATH and keeps the rest unchanged. Thus the quotient complex MATH is MATH-linear and the map is the multiplication by MATH. In particular, MATH fulfills REF for all MATH and taking MATH, we may assume by REF that MATH. If MATH, then there is a meromorphic section MATH of MATH verifying the flatness condition MATH . This implies in particular that MATH has at most logarithmic poles along MATH, which contradicts the condition MATH. On the other hand, MATH for dimension reasons, thus we can apply REF together with REF to obtain the result, after we have replaced MATH by MATH to trivialize the connection at MATH and applied the projection formula to this tensor connection. |
math/9904088 | The assumption that the curvature is vertical implies MATH . Multiplying through by MATH and contracting against MATH we deduce REF. For REF, we must show MATH. From REF, using MATH we reduce to showing MATH. Since MATH has entries MATH, one has MATH . And the residue of an exact form is vanishing. |
math/9904088 | First we show independence of the choice of lifting of MATH. As remarked above, MATH is determined by the local lifting of MATH to MATH, so MATH depends only on that choice. If MATH and MATH are two such local liftings, with MATH, we have MATH for some MATH. It follows immediately that MATH as desired. Next we show independence of the choice of trivializations themselves. Let MATH be a rational function on MATH whose divisor MATH is disjoint from the singular locus of MATH. It will suffice to show that MATH and MATH as trivializations give rise to the same invariant, that is, MATH . Recall we can calculate MATH by choosing MATH a divisor in the linear series MATH compatible with the rigidification MATH and then restricting MATH and taking the norm to MATH. Associated to the trivialization MATH we may take the divisor MATH. It follows that MATH (To get this relation, one could have taken REF as well). On the other hand, since the formula depends only on the local behavior of MATH near MATH, by suitably choosing MATH, we may assume MATH is defined by MATH in a neighborhood of MATH and that MATH modulo some large power of the maximal ideal at the finite set of points where the connection is not given by MATH. We can interpret MATH, so the sum of the residues over all closed points of MATH will vanish in MATH. Thus MATH . Since the connection matrix for the determinant connection is the trace of the connection matrix, the contributions to REF cancel. |
math/9904088 | The usual exact sequence reduces us to showing the condition is equivalent to MATH . Writing as usual MATH where MATH is the reduced divisor with support equal to the support of MATH, we have a commutative square MATH . The map MATH is a quasi-isomorphism if and only if for all MATH the map MATH on the bottom line of the above square is an isomorphism, and this will hold only if the top line is. In particular, since MATH has poles of order REF, if all multiplicities are MATH, then MATH is a quasiisomorphism if and only if MATH is an isomorphism. |
math/9904088 | First MATH . Next computing mod the ideal of MATH and so ignoring the tilde, MATH . By our commutation REF MATH . Also terms with no poles (that is, terms not involving MATH) can be ignored. We get MATH which is what we want. Note the right hand equality holds because we are computing mod forms regular along MATH, and MATH mod regular forms by the cocycle REF . It remains to show that our cocyle REF does not depend on the choice of the cocycles MATH. If MATH is replaced by MATH, then the new REF differs from the old one by the cochain MATH . If MATH is replaced by MATH, then MATH is replaced by MATH, and MATH is replaced by MATH. The commutativity relation implies MATH. Then the new REF differs from the old one by MATH . Using the commutativity relation REF as well, we see that this is this expression is the cochain MATH . |
math/9904088 | Another representative of MATH in MATH is of the shape MATH, thus its derivative in MATH is of the shape MATH since MATH. Then one applies the commutativity of the diagramm MATH . |
math/9904088 | Since, under the integrabiltiy assumption, one has in particular MATH, one can apply REF . One has to compute MATH . We omit the indices since we compute only with one index. Note in the calculations which follow MATH is regular and MATH has poles along MATH. We write MATH to indicate that the polar parts of MATH and MATH coincide. REF implies MATH thus MATH which implies MATH . So taking the trace, one obtains MATH . Now, under the integrability assumption MATH (using REF) one obtains MATH . Now we consider the other term MATH . Choosing a local basis of the bundle MATH, we write MATH as a matrix MATH, and MATH as matrix MATH. Then REF reads MATH and one has MATH . REF implies MATH thus MATH . On the other hand, differentiating REF , one obtains MATH . So taking the trace, one obtains MATH . Thus MATH. |
math/9904088 | We first check injectivity. Let MATH be a local defining equation for MATH. Suppose for some MATH with MATH we had MATH . Multiplying by MATH and taking residue along MATH, it would follow that MATH, that is, MATH. To show surjectivity, write a local section of MATH (here MATH) in the form MATH where MATH does not involve MATH. Replacing MATH by MATH, we can assume MATH . Then MATH . Multiplying by MATH and taking residue along MATH, we see MATH. Since MATH does not involve MATH, it follows that MATH and MATH can be written MATH . Comparing with REF, we have lowered the order of pole by MATH. This process continues until MATH has log poles. |
math/9904088 | Using the lemma, we get a diagram with exact rows MATH . We view this as a diagram of complexes written vertically. Using the standard hypercohomological interpretation of line bundles with connection, this yields an exact sequence MATH . We claim the map MATH above is zero. In the derived category, MATH factors MATH . We have a factorization of MATH: MATH so it suffices to show the map MATH is zero. By NAME theory, the composition MATH is zero, and the map MATH is injective as the complex MATH is quasi-isomorphic to the complex MATH, and in particular starts in degree REF. |
math/9904091 | We proceed by induction on MATH. For MATH, there is nothing to prove. Let us suppose now MATH and let MATH where MATH and MATH. Let MATH then MATH where MATH. If there exist MATH such that MATH, and MATH then, since MATH, it easily follows MATH: thus the lemma is true by induction. Otherwise, if such MATH do not exist, then: MATH . |
math/9904091 | Let MATH and let MATH be a basis of MATH: if MATH, then MATH is a basis of MATH. The weights of MATH are MATH (compare CITE, pag. REF) and since the weights of MATH are given by the sums of couples of different weights of MATH, it easily follows: MATH . Indeed if MATH, then MATH. Let us prove now that MATH does not contain any decomposable bivector, as required. We suppose that there exists MATH, such that MATH; by the previous lemma, we get: MATH where MATH. We want to show that, in this case, there exists a vector of weight MATH in MATH: this contradicts with the fact that MATH. Let MATH, and let MATH be the corresponding endomorphisms of MATH. If we suppose MATH, we have: MATH . Hence if MATH, then MATH. On the other hand it results that MATH, where MATH is a positive integer: this implies that MATH and since MATH, we see that MATH contains a vector of weight MATH. |
math/9904091 | Let MATH. By the NAME criterion CITE, it suffices to show that MATH for any MATH. By the sequence: MATH obtained raising the sequence REF to the MATH-th symmetric power, we see that: MATH for any MATH and MATH. On the other hand by REF , we have the long exact sequence: MATH . This sequence immediately implies that MATH, and since MATH we have that MATH for any MATH, and so MATH is stable. Let us prove now that the condition is necessary. By the sequences: MATH it follows that if MATH, then MATH and so MATH cannot be stable. |
math/9904091 | The proof of this lemma is very similar to the proof of REF. |
math/9904091 | By joining together the sequences REF , we get: MATH . By REF and by the fact that MATH, the last sequence is the minimal resolution of MATH: hence MATH is directly defined by this resolution. |
math/9904091 | By the sequence MATH and since MATH we get MATH hence the lemma is proven. |
math/9904091 | It follows from the definition of MATH, (compare CITE). |
math/9904091 | For brevity's sake, we will write MATH instead of MATH and MATH for MATH. Let also MATH be such that MATH. Let MATH be the sub-variety of the irreducible component of MATH composed by all the quotients of the maps MATH for some weighted bundle MATH and containing the point MATH corresponding to MATH: the morphisms MATH and MATH are canonically defined. A generic fiber of MATH is given by all the cokernels of the morphisms MATH with a fixed MATH, and so, by REF , its dimension is constantly equal (MATH and MATH are fixed) to MATH. Hence, since REF implies that MATH, we get: MATH . Let us study now the morphism MATH: if MATH, then it results MATH and by REF , it follows: MATH . By the sequence: MATH obtained tensoring the dual sequence of REF with MATH, we have that: MATH and so: MATH . Hence: MATH . To prove the proposition it suffices to show that MATH . In fact this implies that MATH, that is, MATH is smooth at the point MATH, and that MATH, that is, MATH is surjective. By the exact sequence: MATH and by the vanishing of MATH and MATH, we have that MATH. Hence by the sequence: MATH and for what we have seen, we get the sequence of cohomology groups: MATH . In particular MATH, as required. |
math/9904095 | Assume first that MATH. The class MATH is represented by the disjoint union MATH. It is clear that the NAME scheme of a disjoint union satisfies MATH . Hence we get MATH. By induction, we get MATH whenever there is a relation MATH for positive integers MATH, MATH and MATH. The corollary follows formally from this. |
math/9904095 | Both sides of this equality are multiplicative in MATH, so it suffices to check it for MATH and MATH. Now the NAME schemes of both MATH and MATH have a MATH-action with isolated fix-points. For such varieties the MATH-genus is given by MATH where the first sum is over all fix-points and MATH is the subspace of the tangent space at MATH generated by eigenvectors for the MATH-action whose eigenvalues have positive weight. In the second sum the MATH are the NAME numbers of the variety. Recall from CITE that the odd NAME numbers for MATH and MATH vanish whereas the even ones are given by MATH and MATH where MATH denotes the number of partitions of the nonnegative integer MATH into MATH parts. This gives MATH and MATH . Now easy calculations using that for any integer MATH we have MATH show that MATH and MATH . On the other hand we have MATH and MATH, and a standard calculation shows that these MATH - genera are related by the exponential expression in the theorem. |
math/9904095 | For a line bundle MATH on MATH let MATH, where MATH and MATH are the two natural maps and where MATH is the projection of MATH onto the MATH-th factor. We will show later that MATH, compare REF . Assume that MATH is a surface such that MATH for some line bundle MATH. Then we claim that the same is true for the NAME scheme MATH. Indeed, we have MATH, and MATH is a MATH-th root of MATH. This shows the theorem for surfaces having a MATH-th root of the canonical line bundle. The formula in the theorem being multiplicative, it will be sufficient to find two independent surfaces having this property. Indeed, a KREF surface and a product of two curves of the same genus MATH such that MATH will do. |
math/9904095 | Let MATH. Then MATH . |
math/9904095 | As MATH is a birational proper morphism of normal varieties, it follows that MATH. MATH has rational singularities, as the quotient of a smooth variety by a finite group (see REF ). Therefore its resolution MATH satifies MATH. |
math/9904095 | Let MATH be a locally free sheaf on MATH. Apply the functor MATH to the exact sequence REF and observe that MATH that is, MATH and similarly MATH. Using MATH we get MATH that is, MATH. |
math/9904095 | We have the following isomorphisms MATH . Here MATH are the higher derived functors of the composite functor MATH. For the first isomorphism see for example, CITE. The last equality is a consequence from the spectral sequence MATH and the observation that the sheaves MATH are supported on the universal family MATH, which is finite over MATH so that all higher direct images vanish. Moreover, MATH and, by MATH, MATH also MATH . Hence we get MATH as an identity in MATH. |
math/9904095 | We have MATH . Now use REF to replace MATH by MATH, where MATH denotes the diagonal. This yields MATH . The two last summands can be simplified as follows: MATH and similarly MATH, since MATH. Finally, MATH, which follows for example, by REF . |
math/9904095 | The morphism MATH is generically finite of degree MATH, so that MATH . Because of an index shift resulting from the insertion of the additional factor MATH between MATH and MATH we have MATH . Using REF we get MATH . And finally by REF MATH . It follows that there are polynomials MATH depending only on MATH, in the NAME classes of the sheaves MATH such that MATH . As we are only trying to prove a general structure result we make no attempt to derive from the above recursion relations for the classes in the MATH-groups more explicit formulae for the dependence of MATH on MATH. Now, according to REF , the last integral equals: MATH . The integrand in this expression is the polynomial MATH. |
math/9904095 | Suppose we are given a polynomial MATH in the NAME classes of MATH. Applying the proposition repeatedly, we may write MATH for some polynomial MATH, which depends only on MATH, in the NAME classes of sheaves on MATH of the form MATH and MATH. Any such expression MATH can be universally reduced to a polynomial expression of integrals of polynomials in the NAME classes of MATH (to see this for the NAME classes of MATH one applies NAME without denominators CITE). This finishes the proof. |
math/9904095 | The proof goes along similar lines as that of REF . The result immediately follows from a modified version of REF : We now allow MATH to be a polynomial in the MATH and the NAME classes of the MATH on MATH, and get MATH to be a polynomial in the MATH and the NAME classes of the analogously defined bundles on MATH. To prove this use the recursion relation REF and the formula MATH to obtain MATH where the MATH are universal polynomials in the MATH and the NAME classes of MATH . Then we again use REF . |
math/9904095 | Let MATH, and let MATH be the map MATH. The images of the five elements MATH, MATH, MATH, MATH, and MATH under MATH are the linearly independent vectors MATH, MATH, MATH, MATH, and MATH. Now, if MATH we may decompose MATH as MATH, where MATH, and get MATH. Moreover, there is a decomposition of the class of the tautological sheaf analogous to the decomposition REF of the NAME scheme: MATH . From the multiplicative behaviour of MATH and MATH we deduce MATH and get MATH . By REF , the function MATH factors through MATH and a map MATH. As the image of MATH is NAME dense in MATH we conclude from REF , that MATH that is, MATH is a linear function. This proves the theorem. |
math/9904095 | Consider the cartesian diagram MATH where MATH is the open subscheme of zero cycles of length MATH whose support consists of at least MATH points and MATH and MATH are the preimages of MATH under MATH and MATH. It is easy to see that MATH is the blow-up of MATH along the (disjoint) diagonals MATH . Let MATH denote the corresponding exceptional divisors, and MATH their sum. The family MATH corresponding to the classifying morphism MATH is the union MATH of the graphs MATH of the MATH-th projection MATH. Projecting the short exact sequence MATH to the base MATH of these families we obtain a short exact sequence MATH of MATH-linearized sheaves. MATH acts on the sheaf in the middle by permutation of the factors. If we take the determinant of the first homomorphism in this sequence, we obtain a short exact sequence MATH of MATH-linearized sheaves with equivariant homomorphisms, where the upper index MATH indicates that the MATH-linearization of MATH is the standard one twisted by the alternating character MATH. (Recall that any two MATH-linearizations of a line bundle on a MATH-scheme differ by a character of the group MATH). Thus we can identify MATH with the MATH-linearized subsheaf MATH, endowed with the alternating linearization. Now MATH and MATH. As MATH is smooth and MATH has codimension MATH, this gives MATH and, similarly, MATH . Here the notation MATH means MATH. Taking dimensions, we get MATH . Now let MATH be an ample line bundle on MATH. By NAME 's construction of the NAME scheme it follows that MATH is very ample for sufficiently ample MATH on MATH. Applying this to MATH for sufficiently large MATH, We conclude that MATH . Evaluating this equation of polynomials in MATH at MATH one finds MATH for all line bundles MATH. Let MATH be an arbitrary line bundle on MATH and let MATH be ample MATH. Then MATH is a MATH-linearized ample line bundle on MATH and descends to an ample line bundle MATH on MATH. For sufficiently large MATH, the line bundles MATH on MATH and MATH on MATH will be very ample. In particular, MATH is globally generated and big. It follows from the NAME vanishing theorem CITE that MATH for all MATH. This gives MATH for all sufficiently large MATH. As both sides are polynomials in MATH, we may take MATH. |
math/9904095 | Using new formal variables MATH and MATH that are related by MATH, MATH, and MATH, so that MATH, and, in the last line of the computation, the NAME formula, we have MATH . This gives MATH and MATH. Now MATH and hence MATH . Differentiating MATH we find MATH . As both MATH and MATH are power series in MATH with constant term REF, we have MATH. Collecting the proven relations we conclude that MATH, MATH and MATH. |
math/9904095 | Let MATH. Then MATH, MATH, MATH. Therefore, by REF , MATH for suitable power series MATH. For the second equality we have used the identities MATH and MATH. It is well-known that MATH. We get by NAME duality MATH . Using MATH and MATH, this gives MATH, MATH, MATH and MATH. To determine MATH and MATH explicitly, let MATH be a KREF-surface. Then by CITE the NAME scheme MATH is an irreducible symplectic complex manifold. There exists a natural quadratic form MATH on MATH (see CITE,CITE,CITE), which on MATH is defined as follows: let MATH be the everywhere non-degenerate holomorphic MATH-form, normalized by MATH. Then for MATH: MATH . Moreover, there exists a universal polynomial MATH such that MATH for all MATH in MATH. For MATH . CITE showed that MATH, MATH, and MATH and MATH are orthogonal with respect to MATH; thus MATH. The polynomial MATH is determined by the formula of REF MATH applied to sufficiently many MATH on MATH with distinct values of MATH: MATH . As MATH, we get MATH. It follows from REF that we can identify MATH and MATH. Finally, the computations of REF show that the equation MATH holds for all surfaces, if we restrict to small ranks MATH. This implies that MATH for MATH. |
math/9904095 | REF follows from REF by putting MATH. In order to prove REF consider the cartesian diagram for MATH where MATH is the NAME morphism and MATH is the image of the universal family MATH in MATH. We denote by MATH, MATH the restriction of the projection. It is easy to see that there is an isomorphism MATH which identifies MATH with the projection to the second factor. The maps MATH and MATH are finite. By REF the higher direct images MATH vanish. An easy spectral sequence argument shows that then the higher direct image sheaves MATH must vanish as well. Moreover, MATH is a birational morphism of integral varities and MATH is normal. This shows that MATH. In particular, for any locally free sheaf MATH on MATH one has MATH. By definition, MATH. As MATH is finite, we get MATH for all MATH. By the NAME formula this gives MATH . By REF , MATH. Hence MATH . The second factor on the right hand side was computed in CITE and equals MATH . |
math/9904104 | In order to make sense of this statement, we will first review both the definition of a holomorphic vertex algebra, and the definition of an algebra in a category. A holomorphic vertex algebra is a vertex algebra without singularities. In other words, the vertex operator of a holomorphic vertex algebra is a map: MATH . It follows that the locality axiom reduces to the statement that products of vertex operators commute. The collection of holomorphic vertex algebras is given the structure of a category by defining a morphism in the category of holomorphic vertex algebras to be a map of complex vector spaces taking vacuum to vacuum, commuting with multiplication, and respecting the actions of the infinitesimal translation operators. A commutative (associative) algebra in a symmetric tensor category consists of an object MATH, a multiplication map MATH which is invariant under the symmetry action for the tensor product, and a unit for the multiplication MATH satisfying the usual axioms for associativity and unit (where MATH is the unit for the tensor product). For the category of representations of the NAME algebra MATH, the unit for multiplication is MATH. and the multiplication map MATH is MATH-invariant. Given any such algebra in this category of representations, where the multiplication map MATH is MATH-invariant, we form a holomorphic vertex algebra as follows. We begin by taking MATH as the infinitesimal translation operator, and take the vacuum vector to be MATH. Then by considering MATH in the extended representation, we have a map MATH which satisfies MATH . From this we define a vertex operator on MATH to be MATH . Checking that this satisfies the axioms for a holomorphic vertex algebra, we see immediately that the vacuum axioms are satisfied because MATH acts trivially on MATH, and because MATH for all MATH. Writing out the translation covariance axiom, we use the MATH-invariance of MATH to give: MATH . And the locality axiom follows from the commutativity of MATH. Notice that MATH. Similarly, given any vertex algebra, we can easily define an algebra in the category of of representations of the NAME algebra MATH by taking MATH for any MATH, and MATH. The unit axioms follow from the properties of the vacuum, and associativity follows from locality. The translation covariance axiom says that the multiplication is MATH-invariant. And, locality axiom acting on the vacuum says that this multiplication is commutative. If MATH is a vertex algebra, and the corresponding MATH-invariant algebra is MATH, then from this algebra, we get back the vertex algebra, MATH. If MATH for MATH, then MATH so setting MATH, we have MATH and so MATH. Similarly, starting with a MATH-invariant algebra, MATH, the process of creating a holomorphic vertex algebra and then mapping back to a MATH-invariant algebra clearly maps MATH to itself. Finally, a morphism in the category of holomorphic can be seen to correspond exactly to a morphism of algebras in the category of MATH-modules. Because this pairing of objects in each category is completely natural we have an isomorphism of categories and so our claim is proved. |
math/9904104 | This is clear from the action defined in REF and the coassociativity of MATH. |
math/9904104 | The MATH-invariance of the singular multilinear maps allows the action of MATH on the domain, MATH, to carry over to an action on the domain of MATH as described in REF. Hence the action of MATH on MATH passes through to an action on MATH exactly when MATH is MATH-invariant. |
math/9904104 | The proof follows immediately from the fact that the evaluation of MATH on MATH gives the same result when carried out by either first evaluating MATH, or by first evaluating MATH or by evaluating both together. |
math/9904104 | We shall only prove that the composition works for the classical vertex group. For the general theory, see CITE. Since composition is pointwise, the composite REF factors through both REF, and so we can focus on those compositions. In order to prove that the composites map into the pullback, we shall first prove that they map into each of the pullback objects (the singular maps given in REF). Taking the composite in REF , we see that it can be considered a multi map as in REF through the following natural map: MATH . Similarly we can can see that it gives a multi map as in REF through evaluating MATH first. From the lemma, we know that these two ways of evaluating are equal, so the composite must factor through the pullback. |
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